Time Dilation: Why Do Observers Measure Different Times?

[Gavroy]

hi

i have some troubles to understand something properly:

if oberserver A is not moving in his inertial frame and there is some observer B moving relatively to oberserver A with velocity V, then this observer B will measure a different(shorter) time from the one that is measured by observer A.

but one could also say that from B's inertial frame that B is not moving and he says: hey, observer A is moving with velocity -V, then the equation that gives me the time dilation would be the same, which would mean, that in this case observer A would measure a shorter time from the one that is measured by oberserver B, which seems to be a logical contradiction, where am i wrong?

 

[Mentz114]

Let's say that A uses coordinates (t,x) to measure things he is comoving with, and B uses coordinates (T,X). If A calculates T in terms of t, he gets dilation of T. If B calculates t in terms of T he gets dilation of t. But on its own it doesn't mean much.

However, if A and B use Doppler radar to measure each others clocks then

if they are separating they both see the others clock run slowly ( red spectral shift)
if they are approaching they both see the others clock running faster ( blue spectral shift)

It's probably best to concentrate on the Doppler because that's what actually happens, whereas the coordinate change isn't.

 

[ghwellsjr]

However, if A and B use Doppler radar to measure each others clocks the...

Good answer, except that it's just plain old Doppler (1-way) not Doppler radar (2-way).

 

[Mentz114]

Good answer, except that it's just plain old Doppler (1-way) not Doppler radar (2-way).

Thanks. You're right about the Doppler.

Just to clarify for the OP - if light is emitted by A or B and received by the other, it will show the spectral shifts. The radar will measure the distance between A and B but that's another tale.

 

[Naty1]

ghwell:

...it's just plain old Doppler

wow, how did you ever notice THAT?? I just skipped right past it...good catch.

 

[ghwellsjr]

ghwell:

...it's just plain old Doppler

wow, how did you ever notice THAT?? I just skipped right past it...good catch.

I suppose because I'm really sensitive to the difference between one-way and two-way things in relativity (such as the speed of light).

 

[stevendaryl]

I think using Doppler shift to illustrate twin paradox effects is helpful, but the problem with it--it seems to me--is that you have to use time dilation in order to derive the proper form for relativistic Doppler shift.

There is, though, a heuristic argument for the relativistic Doppler shift that is possibly more direct than the whole derivation of the Lorentz transformations.

Assume that in some frame F, light travels isotropically at speed c. Let there be one clock, A, that is at rest in frame F, that sends out light signals at a characteristic rate of one per second. Let a second clock, B be traveling at speed v relative to F. Let it be sending signals at a characteristic rate of r (to be determined). r is measured relative to frame F.

Now, we can compute two different ratios:
R_AB = (rate at which signals sent by A arrive at B)/(rate at which B sends signals)
R_BA = (rate at which signals sent by B arrive at A)/(rate at which A sends signals)

R_AB = (1-(v/c))/r
R_BA = r/(1+(v/c))

Now, let's invoke the relativity principle: Assume that the situation between clocks A and B are symmetrical. That's only possible if R_AB = R_BA, which implies
(1-(v/c))/r = r/(1+(v/c))

which implies r = √(1-(v/c)²)

With that choice for r, we have
R_AB = R_BA = √((c-v)/(c+v))

So the relativistic Doppler shift formula is the only choice that satisfies the relativity principle and the isotropy of the speed of light.